C - Xenia and Ringroad

Problem description

Xenia lives in a city that has n houses built along the main ringroad. The ringroad houses are numbered 1 through n in the clockwise order. The ringroad traffic is one way and also is clockwise.

Xenia has recently moved into the ringroad house number 1. As a result, she's got mthings to do. In order to complete the i-th task, she needs to be in the house number a_i and complete all tasks with numbers less than i. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time.

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵). The second line contains m integers a₁, a₂, ..., a_m (1 ≤ a_i ≤ n). Note that Xenia can have multiple consecutive tasks in one house.

Output

Print a single integer — the time Xenia needs to complete all tasks.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

4 3
3 2 3

Output

Input

4 3
2 3 3

Output

Note

In the first test example the sequence of Xenia's moves along the ringroad looks as follows: 1 → 2 → 3 → 4 → 1 → 2 → 3. This is optimal sequence. So, she needs 6 time units.

解题思路：如果后一个数b比前一个数a大，就累加它们的差值b-a；如果后一个数b比前一个数a小，就先循环一圈到后一个数b，此时累加数为n-a+b，注意：因为和最大值可能为1e10（11位）已经爆int，所以ans要用long long。水过！

AC代码：

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long LL;

 int n,m;LL ans,a,b;

 int main(){

     cin>>n>>m;ans=,a=;

     for(int i=;i<=m;++i){

         cin>>b;

         if(b>=a)ans+=b-a;

         else ans+=n-a+b;

         a=b;

     }

     cout<<ans<<endl;

     return ;

 }

巴特西

C - Xenia and Ringroad

Problem description

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Output

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Output

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